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Book X]. with the figure LQ, and the straight line CG with MQ, and
the point G with the point Q. since therefore all the planes and sides of the folid figure AG coincide with the planes and sides of the folid figure KQ, AG is equal and similar to KQ. and in the same manner, any other solid figures whatever contained by the same number of equal and similar planes, alike situated, and having none of their folid angles contained by more than three plane angles, - may be proved to be equal and similar to one another. Q. E. D.
IF a folid, be contained by fix planes, two and ciro of
which are parallel; the opposite planes are fimilar and equal parallelograms.
Let the solid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE. its opposite planes are similar and equal parallelograms.
Because the two parallel planes, BG, CE are cut by the plane 3.16.11. $C, their common sections AB, CD are parallela. again, because
the two parallel planes BF, AE arc cut by the plane AC, their com-
D another and are not in the same plane b. 10. 11. with the other two; wherefore they contain equal angles b; the
angle ABH is therefore coral to the angle DCF. and because AB, BH are cqual to DC, CF, and the angle ABH equal to the angle DCF, thercfore the base AH is equal to the base DF, and the tri
angle ABH to the triangle DCF. and the parallelogram BG is d. 34. 1. double d of the triangle ABH, and the parallelogram CE double of
the triangle DCF; therefore the parallelogram BG is equal and fimilar to the parallelogram CE. in the fame manner, it may be proved that the parallelogram AC is equal and similar to the parallelo
10. 4. I.
gram GF, and the parallelogram AE to BF. Therefore if a fo- Book XI. lid, &c. Q. E. D.
F a solid parallelepiped be cut by a plane parallel to See N.
iwo of its opposite planes; it divides the whole into two solids, the base of one of which shall be to the base of the other, as the one folid is to the other.
Let the folid parallelepiped ABCD be cut by the plane EV which is parallel to the opposite planes AR, HD, and divides the whole into the two solids ABFV, EGCD; as the base AEFY of the first is to the base EHCF of the other, so is the folid ABFV to the folid EGCD.
Produce AH both ways, and take any number of straight lines HM, MN each equal to EH, and any number AK, KL each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the solids LP, KR, HU, MT. then because he straight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF are equal. a. 36. 1. X B G
1 P R
and likewise the parallelograms KX, KB, AG”; as also b thc pa- b. 24. 11. rallelograms LZ, KP, AR, because they are opposite planes. for the same reason, the parallelograms EC, HQ; MS are equal“; and the parallelograms HG, HI, IN, as also - HD, MU, NT. therefore three planes of the folid LP, are equal and similar to three planes of the solid KR, as also to three planes of the solid AV. but the three planes opposite to these three are equal and similar to them in the several solids, and none of their solid angles are contained by more than three plane angles. therefore the three solids LP, KR, AV are equal to one another. for the same reason, the c. C. 15. three solids ED, HU, MT are equal to one another. therefore what
Book XI. multiple soever the base LF is of the base AF, the same multiple is mthe folid LV of the folid AV. for the same reafon, whatever mul,
tiple the base NF is of the base HF, the same multiple is the solid
NV of the solid ED. and if the base LF be equal to the base NF, c. C. u. the folid LV is equal to the solid NV; and if the base LF be
greater than the base NF, the folid LV is greater than the folid NV;
bases AF, FH, and the two folids AV, ED, and of the base AF and folid AV, the base LF and folid LV are any equimultiples whatever; and of the base FH and solid ED, the base FN and solid NV are any equimultiples whatever ; and it has been proved, that if the base
LF is greater than the base FN, the solid LV is greater than the fod. 5. Def. s. lid NV, and if equal, equal; and if less, less. Therefore d as the
base AF is to the base FH, fo is the folid AV to the folid ED. Wherefore if a solid, &c. Q. E. D.
T a given point in a given straight line, to make a
solid angle equal to a given solid angle contained by three plane angles.
Let AB be a given straight line, A a given point in it, and Da given solid angle contained by the three plane angles EDC, EDF, FDC. it is required to make at the point A in the straight line AB a solid angle equal to the solid angle D.
In the straight line DF take any point F, from which draw . FG
perpendicular to the plane EDC, meeting that plane in G; join DG, b. 23. 1. and at the point A in the straight line AB make b the angle BAL e
qual to the angle EDC, and in the plane BAL make the angle BAK equal to the angle EDG;, then make AK equal to DG, and from
f. 8. 1.
the point K erectKH at right angles to the plane BAL; and make Book XI. KH equal to GF, and join AH. then the solid angle at A which is m contained by the three plane angles BAL, BAH, HAL is equal to C. 19, 18. the solid angle at D contained by the three plane angles EDC, EDF, FDC.
Take the equal straight lines AB, DE, and join HB, KB, Fe, GE. and because FG is perpendicular to the plane EDC, it makes right angles d with every straight line meeting it in that plane. d.3.Defil therefore each of the angles FGD, FGE is a right angle. for the same reason, HKA, HKB are right angles. and because KA, AB are equal to GD, DE, each to each, and contain equal angles, therefore the base BK is equal to the base EG. and KH is equal to GF, e. 4. l. and HKB, FGE are right angles, therefore HB is equal to FE. again, because AK, KH are equal to DG, GF, and contain right angles, the base AH is equal to the base DF; and AB is equal to DE; therefore HA, AB are equal to FD, DE, and the base HB is equal to the base FE; there
A fore the angle BAH
D is equal to the angle EDF. for the same reason, the angle HAL is equal to the angle
F qual, and KL, HL, GC, FC be joined, since the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG are, by the construction, equal ; therefore the remaining angle KAL is equal to the remaining angle GDC. and because KA, AL are equal to GD, DC, and contain equal angles, the base KL is equal to the base GC. and KH is equal to GF, so that LK, KH are equal to CG, GF, and they contain right angles; therefore the base HL is equal to the base FC. again, because HA, AL are equal to FD, DC, and the base. HL to the base FC, the angle HAL is equalf to the angle FDC. therefore because the three plane angles BAL, BAH, HAL which contain the solid angle at A, are equal to the three plane angles EDC, EDF, FDC which contain the solid angle at D, each to each, and are situated in the same order; the solid angle at A is equal to g. B. 11. the folid angle at D. Therefore at a given point in a given straight
Book XI. line a solid angle has been made equal to a given solid angle conWtained by three plane angles. Which was to be done.
PROP. XXVII. PROB.
O describe from a given straight line a solid paralle
lepiped similar, and similarly situated to one given.
Let AB be the given straight line, and CD the given solid parallelepiped. It is required from AB to describe a solid parallelepiped similar, and similarly situated to CD.
At the point A of the given straight line AB make a solid angle equal to the solid angle at C, and let BAK, KAH, HAB be the three plane angles which contain it, so that BAK be e
qual to the angle ECG, and KAH to GCF, and HAB to FCE. b. 12. 6. and as EC to CG, so make b BA to AK, and as GC to CF, fo S. 22. s. make b KA to AH; wherefore, ex aequali“, as EC to CF, so is
BA to AH. complete the parallelogram BH, and the solid AL. and
parallelograms of the solid AL are similar to three of the solid CD; ; d. 24. II. and the three opposite ones in each solid are equal d and similar to
these, each to each. also, because the plane angles which contain
the solid angles of the figures are equal, each to each, and situated in e. B. 11. the same order, the solid angles are equal é, each to each. Theref. 11. Def. fore the folid AL is similar f to the solid CD. wherefore from a gi
ven straight line AB a solid parallelepiped AL has been described similar, and similarly situated to the given one CD. Which was to be done.